Good approach. Thank you

So from 1970 to 1979, the cost went up from $2,000 to $3,600 at a fixed percentage.

Letting \(i\) be the fixed percentage, this can be written as:

\(2000(1 + i)^9 = 3600\)

Then, from 1979 to 1988, the cost went up again by the same fixed percentage. Let's call the new amount \(x\). This can be written as:

\(3600(1 + i)^9 = x\)

Now lets put both of these equation on top of eachother:

\(2000(1 + i)^9 = 3600\)
\(3600(1 + i)^9 = x\)

We can divide the top equation by the bottom equation to get rid of that \((1 + i)^9\). If that's not a technique you're comfortable with, you can also just isolate \((1 + i)^9\), and then set up the proportion:

\(\frac{2000}{3600}=\frac{3600}{x}\)

Simplifying:

\(\frac{5}{9}=\frac{3600}{x}\)

\(5x=3600*9\)
\(x=720*9\)
\(x=700*9 + 20*9\)
\(x=6300 + 180\)
\(x=6480\)

And there's answer E

Its 3600/2000 not 3600/1200

since there is a percentage increase. 3600 - 2000 / 2000 * 100. = 80% change every 9 years.

Similarly getting a percent change from the year 1979 to 1988. there is a gap of 9 years. 80% of 3600 = 2880. which when added to 3600 + 2800 gives 6480.

I always follow one rule in GRE. if the questions seem complicated and are taking more than a minute to finish. You are doing it wrong